98
Chao
ref
2
pix=,
=
p,(t).
An additional condition must not be set in the area
of fluid flow-out. Therefore, the movement domain
is
not known in
advance when there
is
a transition
to
Lagrange coordinates
(5,
t)
The deduction of a priori bounds for density in this case gets
essentially difficult. Investigation of a flow problem began in
L55]
on the base of the Burgers model system. S.Ya. Belov
[30]
has recently enhanced these results, however, a general system
hasn't been studied yet
(1.1
1.
4.
A
new formulation of a problem on gas movement caused by the
movement of a solid wall (piston problem) has been suggested in
the work
[55].
On the one hand a problem with the condition of
adhesion to the wall
is
valid. On the other hand, actually,
if
a
piston
is
pulled out of gas with a rather large velocity, the gas
takes off the piston. !'hen there
is
a lack of viscosity in gas
there exist some examples
of
precise solutions of such problems
L117].
Therefore the following set up of a problem on a piston for
system
(1.1)
is
more physically based.
Let
gas at an initial moment
occupy a semi-axis
x<O
and the right boundary move by the law
x
=
e(t). Gas may move
so
that an extreme right particle
is
situa-
ted on the wall z(t)
the wall
is
possible when there arises a free boundary x
=
y(t)
on which correlations hold
and that's why ~l~=~(~)
=
dz/dt.
A
take off
It
is
reasonable to consider that the piston takes off,
if
strain
u=
p
-u-
z:
reaches the minimum value, zero, for instance, cor-
responding to the strain on a free boundary.
If
the piston doesn't
take off, strain on the boundary
is
non-negative. Besides, the
condition of non-fluidity through a solid wall can be formulated
in the form of inequality on
a
sought function y(t):
y(t)
<-
z(t).
In a general form the boundary value problem
is
lowing way:
it
is
required to find the solution
the domain-w
<
x
<
y(t),
t
E
[0,
T]
with an
x
=
y(t)
on which conditions are satisfied
set up
in
the fol-
of system
(1.1)
in
known boundary
(9.3)
The last equality means that either condition
Y=Z
Or
0.
Q=
0
condition on a free boundary are satisfied. Boundary value condi-
tions
(9.3)
contain inequalities, and the problem becomes one-
-sided. After the transition to Lagrange coordinates the boundary
x
=
y(t) becomes known and then for system
(1.10)
there appears a
roblem with boundary conditions for
x
=
0
of the following
form
a
x
is
a Lagrange variable):